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1.
a. Ta có: \(A=2^{300}=2^{3.100}=\left(2^3\right)^{100}=8^{100}\)
\(B=3^{200}=3^{2.100}=\left(3^2\right)^{100}=9^{100}\)
Mà \(8^{100}< 9^{100}\)
\(\Rightarrow A< B\)
b. Ta có: \(A=2^{332}< 2^{333}=2^{3.111}=\left(2^3\right)^{111}=8^{111}\)
\(B=3^{223}>3^{222}=3^{2.111}=\left(3^2\right)^{111}=9^{111}\)
Mà \(8^{111}< 9^{111}\)
\(\Rightarrow A< B\)
c. Ta có: \(A=2^{91}=2^{13.7}=\left(2^{13}\right)^7=8192^7\)
\(B=5^{35}=5^{5.7}=\left(5^5\right)^7=3125^7\)
Mà \(8192^7>3125^7\)
\(\Rightarrow A>B\)
Câu 2:
a: =>(x-6)(x-7)=0
=>x=6 hoặc x=7
b: =>\(x^8\left(x^2-25\right)=0\)
\(\Leftrightarrow x^8\left(x-5\right)\left(x+5\right)=0\)
hay \(x\in\left\{0;5;-5\right\}\)
a)(357.252.714):510=(57.77.54.714):510=(511.721):510=5.721
b)=215.520.212.54 /22.52=227.524/22.52=225.522
7: =6-7+4-3=0
8: =8x5-9x4+4x6=40-36+24=28
3: =231+26-209-26=22
\(3^3\cdot5^3-20\cdot\left\{300-\left[546-2^3\cdot\left(7^8-7^6+7^0\right)\right]\right\}\)
\(=3^3\cdot5^3-20\cdot\left\{300-\left[546-2^3\cdot5647153\right]\right\}\)
\(=3^3\cdot5^3-20\cdot\left\{300-\left[546-45177224\right]\right\}\)
\(=3^3\cdot5^3-20\cdot\left\{300--45176678\right\}\)
\(=3^3\cdot5^3-20\cdot45176978\)
\(=3375-903539560\)
\(=-903536185\)
\(626500:\left\{50^2:\left[178-4\cdot\left(35-21:3\right)\right]\right\}\)
\(=626500:\left\{50^2:\left[178-4\cdot\left(35-7\right)\right]\right\}\)
\(=626500:\left\{50^2:\left[178-4\cdot28\right]\right\}\)
\(=626500:\left\{50^2:\left[178-112\right]\right\}\)
\(=626500:\left\{50^2:66\right\}\)
\(=626500:\frac{1250}{33}\)
\(=\frac{82698}{5}\)
CHUC BAN HOC TOT >.<
\(\left(\frac{3}{7}+\frac{1}{2}\right)^2=\left(\frac{13}{14}\right)^2=\frac{169}{196}\)
\(\frac{5^4\cdot20^4}{25^5\cdot4^5}=\frac{5^4\cdot4^4\cdot5^4}{5^5\cdot5^5.4^5}=\frac{1}{100}\)
\(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6=\left(\frac{3}{7}\right)^{21}:\left[\left(\frac{3}{7}\right)^2\right]^6=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}=\left(\frac{3}{7}\right)^9\)
\(3-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2\)
\(=3-1+\frac{1}{4}:2=3-1+\frac{1}{8}=\frac{17}{8}\)
1. A - B = 40+ 3/8 + 7/82 + 5/83 + 32/85 - (24/82 + 40+ 5/82 + 40/84 + 5/84 )
= 40.85/85 + 3.84/85 + 7.83/85 + 5.82/85 + 32/85 - 24.83/85 - 40.85/85 - 5.83/85 - 40.8/85 - 5.8/85
= 40.85/85 + 24.83/85 + 7.83/85 + 5.82/85 + 32/85 - 24.83/85 - 40.85/85 - 5.83/85 - 40.8/85 - 5.8/85
= 7.83/85 + 5.82/85 + 32/85 - 5.83/85 - 40.8/85 - 5.8/85
= 7.83/85 + 5.82/85 -8/85 - 5.83/85 - 40.8/85
= 2.83/85 + 5.82/85 - 40.8/85 - 8/85
= 2.83/85 + 40.8/85 - 40.8/85 - 8/85
= 2.83/85 - 8/85 > 0
Vay A > B
= (325 +320)(240 + 735) .0 = 0